JEE Main 2023 — Mathematics Algebra

Let us consider four positive consecutive terms of a G.P. as follow:

$$a,ar,a{r}^{2},a{r}^{3}(a,r>0)$$a,ar,ar2,ar3(a,r>0)

Product $=$=${a}^{4}{r}^{6}=1296$a4r6=1296

$$\Rightarrow {a}^{2}{r}^{3}=36$$⇒a2r3=36

$$\Rightarrow a=\frac{6}{{r}^{3/2}}$$⇒a=r3/26​

Sum $=$=$a+ar+a{r}^{2}+a{r}^{3}=126$a+ar+ar2+ar3=126

$$\Rightarrow \frac{1}{{r}^{3/2}}+\frac{r}{{r}^{3/2}}+\frac{{r}^{2}}{{r}^{3/2}}+\frac{{r}^{3}}{{r}^{3/2}}=\frac{126}{6}$$⇒r3/21​+r3/2r​+r3/2r2​+r3/2r3​=6126​

$$\Rightarrow ({r}^{-3/2}+{r}^{3/2})+({r}^{1/2}+{r}^{-1/2})=21$$⇒(r−3/2+r3/2)+(r1/2+r−1/2)=21

Let ${r}^{1/2}+{r}^{-1/2}=A$r1/2+r−1/2=A

$$\Rightarrow {r}^{-3/2}+{r}^{3/2}=({r}^{\frac{1}{2}}+{r}^{\frac{-1}{2}}{)}^{3}-3({r}^{\frac{1}{2}}+{r}^{\frac{-1}{2}})$$⇒r−3/2+r3/2=(r21​+r2−1​)3−3(r21​+r2−1​)

$$={A}^{3}-3A$$=A3−3A

Therefore,

$${A}^{3}-3A+A=21$$A3−3A+A=21

$$\Rightarrow {A}^{3}-2A=21$$⇒A3−2A=21

$$\Rightarrow A=3$$⇒A=3

$$\therefore \sqrt{r}+\frac{1}{\sqrt{r}}=3$$∴r​+r​1​=3

$$\Rightarrow r+1=3\sqrt{r}$$⇒r+1=3r​

On squaring both sides:

$${r}^{2}+2r+1=9r$$r2+2r+1=9r

$$\Rightarrow {r}^{2}-7r+1=0$$⇒r2−7r+1=0

Here, sum of roots$=7$=7